Question

A 1.4-µC point charge is placed between the plates of a parallel plate capacitor. The charge experiences a force of 0.38 N. What is the magnitude σ of the charge density on either plate of the capacitor?

Answer 1

Answer:

2.4 * 10^-6 C/m²

Explanation:

The Force experimented by a particle in an electric field will be equal to:

F = E*q

Where E is the magnitude of the electric field and q is the charge of the particle.

The Electric field caused by a parallel plate capacitor will be:

E = σ/ε0

Where σ is the charge density of the plates of the capacitor and ε0 is the Vacuum permittivity constant.

We replace this expression in the expression for force and issolate σ and get that:

σ = F*ε0/q = 0.38 N * 8.854 * 10^-12 F/m / (1.4*10^-6C) = 2.4 * 10^-6 C/m²